【Rotate List】cpp

题目

Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (!head || !(head->next)) return head;
        ListNode *p = head;
        int len = 1;
        for (;p->next;++len,p=p->next){}
        ListNode *end = p;
        k = k % len;
        p = head;
        for (size_t i = 0; i < len-k-1; ++i)
        {
            p = p->next;
        }
        end->next = head;
        head = p->next;
        p->next = NULL;
        return head;
    }
};

Tips:

思路很简单,需要注意的是对指针的操作。

=====================================

第二次过这道题:

(1)没有考虑k比ListNodes长度大的情况

(2)没考虑k加上双指针的边界情况没有考虑完全,所以几次都没有AC。

完备的思路应该是:先求出来ListNodes的长度,k%len就是真正要rotate的元素。搞清楚这之后,再利用快慢指针常规思路就可以解出来了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
            if ( !head || !head->next ) return head;
            int len = 1;
            for ( ListNode* p = head; p->next; ++len, p=p->next){}
            k = k % len;
            ListNode dummpy(-1);
            dummpy.next = head;
            ListNode* p1 = head;
            ListNode* p2 = head;
            for ( int i=0; inext;
            while ( p2->next )
            {
                p1 = p1->next;
                p2 = p2->next;
            }
            p2->next = head;
            dummpy.next = p1->next;
            p1->next = NULL;
            return dummpy.next;
    }
};

 

转载于:https://www.cnblogs.com/xbf9xbf/p/4467738.html

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